$a)\left(\frac{3}{4}+\frac{2}{3}\right)\cdot 6=\left(\frac{9}{12}+\frac{8}{12}\right)\cdot 6=$

$=\frac{17}{{\sout{12}}^2}\cdot {\sout{6}}^1=\frac{17}{2}=8\frac{1}{2}$  

$b)1\frac{5}{6}-2\cdot \frac{3}{4}=\frac{11}{6}-{\sout{2}}^1\cdot \frac{3}{{\sout{4}}^2}=\frac{11}{6}-\frac{3}{2}=\frac{11}{6}-\frac{9}{6}=\frac{2}{6}=\frac{1}{3}$

$c)1\frac{7}{10}\cdot 10+3\cdot 1\frac{1}{12}=\frac{17}{{\sout{10}}^1}\cdot {\sout{10}}^1+{\sout{3}}^1\cdot \frac{13}{{\sout{12}}^4}=17+\frac{13}{4}=17+3\frac{1}{4}=20\frac{1}{4}$